civil engineering formulas pocket guide kingdwarf

1. TLFeBOOK

2. 40816_FM_pi-xvii 10/22/01 12:37 PM Page i 40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.iCIVIL ENGINEERINGFORMULASTLFeBOOK 3. This page intentionally left blank.TLFeBOOK 4. 40816_FM_pi-xvii10/22/0112:37 PM Page iii40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.iii CIVILENGINEERING FORMULAS Tyler G. Hicks, P.E.International Engineering AssociatesMember: American Society of Mechanical EngineersUnited States Naval Institute McGRAW-HILL New York Chicago San Francisco Lisbon LondonMadrid Mexico City Milan New Delhi San JuanSeoul Singapore Sydney TorontoTLFeBOOK 5. McGraw-HillabcCopyright 2002 by The McGraw-Hill Companies. All rights reserved. Manufactured in the UnitedStates of America. 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Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental, special, punitive, consequential orsimilar damages that result from the use of or inability to use the work, even if any of them hasbeen advised of the possibility of such damages. This limitation of liability shall apply to anyclaim or cause whatsoever whether such claim or cause arises in contract, tort or otherwise.DOI: 10.1036/0071395423 TLFeBOOK 6. 40816_FM_pi-xvii 10/22/01 12:37 PM Page v40816 HICKS Mcghp FM Second Passbcj 7/19/01 p.v CONTENTS9 .tPreface xiiinAcknowledgments xv-How to Use This Book xviiChapter 1. Conversion Factors for CivilEngineering Practice1Chapter 2. Beam Formulas 15Continuous Beams / 16Ultimate Strength of Continuous Beams / 53Beams of Uniform Strength / 63Safe Loads for Beams of Various Types / 64Rolling and Moving Loads / 79Curved Beams / 82Elastic Lateral Buckling of Beams / 88Combined Axial and Bending Loads / 92Unsymmetrical Bending / 93Eccentric Loading / 94Natural Circular Frequencies and Natural Periods of Vibration of Prismatic Beams / 96TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 7. 40816_FM_pi-xvii 10/22/01 12:37 PMPage vi40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.vi vi CONTENTS Chapter 3. Column Formulas 99TC General Considerations / 100 C Short Columns / 102R Eccentric Loads on Columns / 102 D Column Base Plate Design / 111 U American Institute of Steel Construction Allowable-StressDesign Approach / 113 W Composite Columns / 115 Elastic Flexural Buckling of Columns / 118 U Allowable Design Loads for Aluminum Columns / 121 Ultimate-Strength Design of Concrete Columns / 124 WUW Chapter 4. Piles and Piling Formulas131UW Allowable Loads on Piles / 132 F Laterally Loaded Vertical Piles / 133F Toe Capacity Load / 134S Groups of Piles / 136C Foundation-Stability Analysis / 139S Axial-Load Capacity of Single Piles / 143B Shaft Settlement / 144 L Shaft Resistance to Cohesionless Soil / 145SCC Chapter 5. Concrete Formulas147W Reinforced Concrete / 148 Water/Cementitious Materials Ratio / 148 C Job Mix Concrete Volume / 149 Modulus of Elasticity of Concrete / 150G Tensile Strength of Concrete / 151 S Reinforcing Steel / 151B Continuous Beams and One-Way Slabs / 151 B Design Methods for Beams, Columns, and Other Members / 153 C Properties in the Hardened State / 167 C TLFeBOOK 8. 40816_FM_pi-xvii 10/22/01 12:37 PM Page viii40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.viii viii CONTENTS Compression at Angle to Grain / 220C Recommendations of the Forest Products Laboratory / 221F Compression on Oblique Plane / 223 S Adjustments Factors for Design Values / 224V Fasteners for Wood / 233 Adjustment of Design Values for Connections withFasteners / 236 C Roof Slope to Prevent Ponding / 238 Bending and Axial Tension / 239L Bending and Axial Compression / 240ALA Chapter 7. Surveying Formulas 243LA Units of Measurement / 244 S Theory of Errors / 245 B Measurement of Distance with Tapes / 247 C Vertical Control / 253 B Stadia Surveying / 253 P Photogrammetry / 255 LCWD Chapter 8. Soil and Earthwork Formulas257FC Physical Properties of Soils / 258 N Index Parameters for Soils / 259 P Relationship of Weights and Volumes in Soils / 261 Internal Friction and Cohesion / 263 Vertical Pressures in Soils / 264 Lateral Pressures in Soils, Forces on Retaining Walls / 265C Lateral Pressure of Cohesionless Soils / 266 F Lateral Pressure of Cohesive Soils / 267 Water Pressure / 268 S Lateral Pressure from Surcharge / 268A Stability of Slopes / 269L Bearing Capacity of Soils / 270A Settlement under Foundations / 271 S Soil Compaction Tests / 272H TLFeBOOK 9. 40816_FM_pi-xvii10/22/0112:37 PMPage ix40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.ix CONTENTS ix Compaction Equipment / 275 Formulas for Earthmoving / 276 Scraper Production / 278 Vibration Control in Blasting / 280 Chapter 9. Building and Structures Formulas 283 Load-and-Resistance Factor Design for Shear in Buildings / 284 Allowable-Stress Design for Building Columns / 285 Load-and-Resistance Factor Design for Building Columns / 287 Allowable-Stress Design for Building Beams / 2873Load-and-Resistance Factor Design for Building Beams / 290 Allowable-Stress Design for Shear in Buildings / 295 Stresses in Thin Shells / 297 Bearing Plates / 298 Column Base Plates / 300 Bearing on Milled Surfaces / 301 Plate Girders in Buildings / 302 Load Distribution to Bents and Shear Walls / 304 Combined Axial Compression or Tension and Bending / 306 Webs under Concentrated Loads / 308 Design of Stiffeners under Loads / 3117 Fasteners for Buildings / 312 Composite Construction / 313 Number of Connectors Required for Building Construction / 316 Ponding Considerations in Buildings / 318 Chapter 10. Bridge and Suspension-Cable Formulas321 Shear Strength Design for Bridges / 322 Allowable-Stress Design for Bridge Columns / 323 Load-and-Resistance Factor Design for Bridge Columns / 324 Allowable-Stress Design for Bridge Beams / 325 Stiffeners on Bridge Girders / 327 Hybrid Bridge Girders / 329TLFeBOOK 10. 40816_FM_pi-xvii10/22/01 12:37 PMPage x40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.x x CONTENTS Load-Factor Design for Bridge Beams / 330 F Bearing on Milled Surfaces / 332O Bridge Fasteners / 333W Composite Construction in Highway Bridges / 333 P Number of Connectors in Bridges / 337 T Allowable-Stress Design for Shear in Bridges / 339F Maximum Width/Thickness Ratios for CompressionCElements for Highway Bridges / 341 O Suspension Cables / 341 M General Relations for Suspension Cables / 345 H Cable Systems / 353 N W F Chapter 11. Highway and Road Formulas355P E M Circular Curves / 356 Parabolic Curves / 359 C Highway Curves and Driver Safety / 361 G Highway Alignments / 362 W Structural Numbers for Flexible Pavements / 365 F Transition (Spiral) Curves / 370 E Designing Highway Culverts / 371 V American Iron and Steel Institute (AISI) Design HProcedure / 374 Chapter 12. Hydraulics and Waterworks Formulas 381 Capillary Action / 382 Viscosity / 386 Pressure on Submerged Curved Surfaces / 387 Fundamentals of Fluid Flow / 388 Similitude for Physical Models / 392 Fluid Flow in Pipes / 395 Pressure (Head) Changes Caused by Pipe Size Change / 403 Flow through Orices / 406TLFeBOOK 11. 40816_FM_pi-xvii10/22/01 12:37 PM Page xi40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xi CONTENTS xi Fluid Jets / 409 Orice Discharge into Diverging Conical Tubes / 410 Water Hammer / 412 Pipe Stresses Perpendicular to the Longitudinal Axis / 412 Temperature Expansion of Pipe / 414 Forces Due to Pipe Bends / 414 Culverts / 417 Open-Channel Flow / 420 Mannings Equation for Open Channels / 424 Hydraulic Jump / 425 Nonuniform Flow in Open Channels / 429 Weirs / 436 Flow Over Weirs / 4385Prediction of Sediment-Delivery Rate / 440 Evaporation and Transpiration / 442 Method for Determining Runoff for MinorHydraulic Structures / 443 Computing Rainfall Intensity / 443 Groundwater / 446 Water Flow for Fireghting / 446 Flow from Wells / 447 Economical Sizing of Distribution Piping / 448 Venturi Meter Flow Computation / 448 Hydroelectric Power Generation / 449Index4511TLFeBOOK 12. This page intentionally left blank.TLFeBOOK 13. 40816_FM_pi-xvii 10/22/01 12:37 PM Page xiii40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.xiiiPREFACE This handy book presents more than 2000 needed formulas for civil engineers to help them in the design ofce, in the eld, and on a variety of construction jobs, anywhere in the world. These formulas are also useful to design drafters, structural engineers, bridge engineers, foundation builders, eld engineers, professional-engineer license examination candidates, concrete specialists, timber-structure builders, and students in a variety of civil engineering pursuits. The book presents formulas needed in 12 different spe- cialized branches of civil engineeringbeams and girders, columns, piles and piling, concrete structures, timber engineering, surveying, soils and earthwork, building structures, bridges, suspension cables, highways and roads, and hydraulics and open-channel ow. Key formulas are presented for each of these topics. Each formula is explained so the engineer, drafter, or designer knows how, where, and when to use the formula in professional work. Formula units are given in both the United States Customary System (USCS) and System International (SI). Hence, the text is usable throughout the world. To assist the civil engineer using this material in worldwide engineering practice, a comprehensive tabulation of conversion factors is presented in Chapter 1. In assembling this collection of formulas, the author was guided by experts who recommended the areas of TLFeBOOKCopyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 14. 40816_FM_pi-xvii10/22/0112:37 PM Page xiv40816 HICKS Mcghp FM Second Pass bcj 7/19/01 p.xiv xiv PREFACE greatest need for a handy book of practical and applied civil engineering formulas.Sources for the formulas presented here include the var- A ious regulatory and industry groups in the eld of civil engineering, authors of recognized books on important topics in the eld, drafters, researchers in the eld of civil engineering, and a number of design engineers who work daily in the eld of civil engineering. These sources are cited in the Acknowledgments.When using any of the formulas in this book that may come from an industry or regulatory code, the userM is cautioned to consult the latest version of the code. t Formulas may be changed from one edition of a code to a the next. In a work of this magnitude it is difcult to H include the latest formulas from the numerous constant- a ly changing codes. Hence, the formulas given here are those current at the time of publication of this book.wIn a work this large it is possible that errors may occur. F Hence, the author will be grateful to any user of the bookp who detects an error and calls it to the authors attention.w Just write the author in care of the publisher. The error willa be corrected in the next printing.LIn addition, if a user believes that one or more important N formulas have been left out, the author will be happy toe consider them for inclusion in the next edition of the book. Again, just write him in care of the publisher. p FTyler G. Hicks, P.E. a s f e w p tTLFeBOOK 15. 40816_FM_pi-xvii 10/22/01 12:37 PM Page xv40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.xvl- ACKNOWLEDGMENTS-n-netr Many engineers, professional societies, industry associa- .tions, and governmental agencies helped the author nd ando assemble the thousands of formulas presented in this book.o Hence, the author wishes to acknowledge this help and- assistance.e The authors principal helper, advisor, and contributorwas the late Frederick S. Merritt, P.E., Consulting Engineer. .For many years Fred and the author were editors on com-k panion magazines at The McGraw-Hill Companies. Fred .was an editor on Engineering-News Record, whereas thel author was an editor on Power magazine. Both lived onLong Island and traveled on the same railroad to and fromt New York City, spending many hours together discussingo engineering, publishing, and book authorship. .When the author was approached by the publisher to pre-pare this book, he turned to Fred Merritt for advice and help.Fred delivered, preparing many of the formulas in this bookand giving the author access to many more in Freds exten-sive les and published materials. The author is most grate-ful to Fred for his extensive help, advice, and guidance.Further, the author thanks the many engineering soci-eties, industry associations, and governmental agencies whosework is referred to in this publication. These organizationsprovide the framework for safe design of numerous struc-tures of many different types.TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 16. 40816_FM_pi-xvii10/22/01 12:37 PM Page xvi40816 HICKS Mcghp FMSecond Pass bcj 7/19/01 p.xvi xvi ACKNOWLEDGMENTS The author also thanks Larry Hager, Senior Editor, Pro- fessional Group, The McGraw-Hill Companies, for his excellent guidance and patience during the long preparation of the manuscript for this book. Finally, the author thanks his wife, Mary Shanley Hicks, a publishing professional, who always most willingly offered help and advice when needed. Specic publications consulted during the preparation of this text include: American Association of State Highway and Transportation Ofcials (AASHTO) Standard Speci- cations for Highway Bridges; American Concrete Institute (ACI) Building Code Requirements for Reinforced Con- crete; American Institute of Steel Construction (AISC)T Manual of Steel Construction, Code of Standard Prac-c tice, and Load and Resistance Factor Design Specica-d tions for Structural Steel Buildings; American Railway Engineering Association (AREA) Manual for Railway s Engineering; American Society of Civil Engineersp (ASCE) Ground Water Management; American Water Works Association (AWWA) Water Quality and Treat- n ment. In addition, the author consulted several hundred T civil engineering reference and textbooks dealing with the i topics in the current book. The author is grateful to theS writers of all the publications cited here for the insight they gave him to civil engineering formulas. A number of theseo works are also cited in the text of this book. dccledida TLFeBOOK 17. 40816_FM_pi-xvii 10/22/01 12:37 PM Page xvii 40816 HICKS Mcghp FM Second Passbcj 7/19/01 p.xvii-snHOW TO USEsn ,THIS BOOKfy-e-) The formulas presented in this book are intended for use by- civil engineers in every aspect of their professional work- design, evaluation, construction, repair, etc.y To nd a suitable formula for the situation you face,y start by consulting the index. Every effort has been made tos present a comprehensive listing of all formulas in the book.r Once you nd the formula you seek, read any accompa-- nying text giving background information about the formula.d Then when you understand the formula and its applications,e insert the numerical values for the variables in the formula.e Solve the formula and use the results for the task at hand.y Where a formula may come from a regulatory code,e or where a code exists for the particular work beingdone, be certain to check the latest edition of the appli-cable code to see that the given formula agrees with thecode formula. If it does not agree, be certain to use thelatest code formula available. Remember, as a designengineer you are responsible for the structures you plan,design, and build. Using the latest edition of any govern-ing code is the only sensible way to produce a safe anddependable design that you will be proud to be associ-ated with. Further, you will sleep more peacefully!TLFeBOOK Copyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 18. This page intentionally left blank.TLFeBOOK 19. 40816_01_p1-1410/22/01 12:38 PM Page 1 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CHAPTER 1 CONVERSION FACTORS FOR CIVIL ENGINEERING PRACTICE TLFeBOOKCopyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 20. 40816_01_p1-14 10/22/01 12:38 PM Page 2 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 2 CHAPTER ONE Civil engineers throughout the world accept both the T United States Customary System (USCS) and the System International (SI) units of measure for both applied and theoretical calculations. However, the SI units are much more widely used than those of the USCS. Hence, both the USCS and the SI units are presented for essentially every formula in this book. Thus, the user of the book can apply s the formulas with condence anywhere in the world. c To permit even wider use of this text, this chapter con- p tains the conversion factors needed to switch from one system to the other. For engineers unfamiliar with either p system of units, the author suggests the following steps for f becoming acquainted with the unknown system:k 1. Prepare a list of measurements commonly used in yourgdaily work. 2. Insert, opposite each known unit, the unit from the other ksystem. Table 1.1 shows such a list of USCS units withcorresponding SI units and symbols prepared by a civilengineer who normally uses the USCS. The SI units wshown in Table 1.1 were obtained from Table 1.3 by theengineer. 3. Find, from a table of conversion factors, such as Table 1.3,the value used to convert from USCS to SI units. Insert ceach appropriate value in Table 1.2 from Table 1.3. a 4. Apply the conversion values wherever necessary for theuformulas in this book.c 5. Recognizehere and nowthat the most difcult Faspect of becoming familiar with a new system of meas-durement is becoming comfortable with the names andmagnitudes of the units. Numerical conversion is simple,once you have set up your own conversion table. TLFeBOOK 21. 40816_01_p1-14 10/22/01 12:38 PM Page 340816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS3e TABLE 1.1 Commonly Used USCS and SI Unitsmd Conversion factorh(multiply USCS uniteby this factor toy USCS unitSI unit SI symbolobtain SI unit)y square footsquare meterm2 0.0929cubic foot cubic meter m3 0.2831- pound per- square inchkilopascalkPa6.894r pound forcenewtonNu 4.448r foot poundtorque newton meterNm1.356kip foot kilonewton meterkNm 1.355r gallon perminute liter per secondL/s0.06309r kip per squareh inch megapascalMPa6.89l This table is abbreviated. For a typical engineering practice, an actual tables would be many times this length.e,Be careful, when using formulas containing a numericalt constant, to convert the constant to that for the system youare using. You can, however, use the formula for the USCSe units (when the formula is given in those units) and thenconvert the nal result to the SI equivalent using Table 1.3.t For the few formulas given in SI units, the reverse proce-- dure should be used.d ,TLFeBOOK 22. 40816_01_p1-14 10/22/01 12:38 PMPage 4 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 4 CHAPTER ONE TABLE 1.2 Typical Conversion TableT To convert fromToMultiply by square foot square meter 9.290304 E 02 a foot per second meter per second a squaredsquared 3.048E 01 a cubic footcubic meter2.831685 E 02 a pound per cubic kilogram per cubic inch meter 2.767990 E 04 a gallon per minute liter per second 6.309E 02 pound per square inchkilopascal 6.894757b pound force newton 4.448222b kip per square foot pascal 4.788026 E 04b acre foot per day cubic meter per E 02Bsecond1.427641 acresquare meter 4.046873 E 03B cubic foot percubic meter per second second2.831685 E 02 This table contains only selected values. See the U.S. Department of the Interior Metric Manual, or National Bureau of Standards, The International System of Units (SI), both available from the U.S. Government Printing B Ofce (GPO), for far more comprehensive listings of conversion factors. The E indicates an exponent, as in scientic notation, followed by a positive or negative number, representing the power of 10 by which the given conversion factor is to be multiplied before use. Thus, for the square foot con-B version factor, 9.290304 1/100 0.09290304, the factor to be used to convert square feet to square meters. For a positive exponent, as in convert- ing acres to square meters, multiply by 4.046873 1000 4046.8.Where a conversion factor cannot be found, simply use the dimensional substitution. Thus, to convert pounds per cubic inch to kilograms per cubic meter, nd 1 lb 0.4535924 kg and 1 in3 0.00001638706 m3. Then, B 1 lb/in3 0.4535924 kg/0.00001638706 m3 27,680.01, or 2.768 E 4. TLFeBOOK 23. 40816_01_p1-1410/22/01 12:38 PMPage 5 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001CONVERSION FACTORS5 TABLE 1.3 Factors for Conversion to SI Units of Measurement To convert from To Multiply by2acre foot, acre ft cubic meter, m3 1.233489E 03 acre square meter, m24.046873E 031angstrom, meter, m1.000000* E 102atmosphere, atmpascal, Pa1.013250* E 05 (standard)4atmosphere, atmpascal, Pa9.806650* E 04 (technical2 1 kgf/cm2) barpascal, Pa1.000000* E 05 barrel (for petroleum, cubic meter, m2 1.589873 E 01442 gal) board foot, board ft cubic meter, m3 2.359737E 032 British thermal unit,joule, J1.05587 E 03 Btu, (mean)3 British thermal unit,watt per meter1.442279E 01 Btu (International kelvin, W/(mK)2 Table)in/(h)(ft2) (F) (k, thermal conductivity) British thermal unit,watt, W 2.930711E 01 Btu (Internationale Table)/h British thermal unit,watt per square 5.678263E 00 Btu (International meter kelvin, Table)/(h)(ft2)(F)W/(m2K) (C, thermal conductance) British thermal unit,joule per kilogram, 2.326000* E 03 Btu (International J/kg Table)/lbTLFeBOOK 24. 40816_01_p1-14 10/22/0112:38 PM Page 6 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 6 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of MeasurementT (Continued)( To convert fromToMultiply by British thermal unit, joule per kilogram 4.186800* E 03(Btu (International kelvin, J/(kgK)Table)/(lb)(F)(c, heat capacity) British thermal unit, joule per cubic3.725895E 04dcubic foot, Btumeter, J/m3f(InternationalfTable)/ft3f bushel (U.S.) cubic meter, m33.523907E 02f calorie (mean)joule, J 4.19002 E 00 candela per squarecandela per square 1.550003E 03sinch, cd/in2 meter, cd/m2 s centimeter, cm, ofpascal, Pa 1.33322 E 03mercury (0C) centimeter, cm, ofpascal, Pa 9.80638 E 01swater (4C) chain meter, m 2.011684 E 01 c circular milsquare meter, m2 5.067075 E 10 day second, s8.640000* E 04c day (sidereal)second, s8.616409 E 04 degree (angle)radian, rad1.745329 E 02 c degree Celsiuskelvin, KTK tC 273.15 degree Fahrenheit degree Celsius, C tC (tF 32)/1.8f degree Fahrenheit kelvin, KTK (tF 459.67)/1.8 degree Rankinekelvin, KTK TR /1.8 (F)(h)(ft2)/Btukelvin square1.761102 E 01 f (Internationalmeter per watt, Table) (R, thermalKm2/W f resistance) TLFeBOOK 25. 40816_01_p1-1410/22/0112:38 PM Page 7 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 CONVERSION FACTORS7 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert fromTo Multiply by2 (F)(h)(ft )/(Btu kelvin meter per6.933471E 00 (Internationalwatt, Km/W Table)in) (thermal resistivity) dyne, dyn newton, N 1.000000 E 05 fathommeter, m1.828804E 00 foot, ftmeter, m3.048000 E 01 foot, ft (U.S. survey)meter, m3.048006E 01 foot, ft, of waterpascal, Pa2.98898 E 03(39.2F) (pressure) square foot, ft2square meter, m29.290304 E 02 square foot per hour, square meter per2.580640 E 05ft2/h (thermal second, m2/sdiffusivity) square foot per square meter per9.290304 E 02second, ft2/ssecond, m2/s cubic foot, ft3 (volume cubic meter, m3 2.831685E 02or section modulus) cubic foot per minute,cubic meter per 4.719474E 04ft3/minsecond, m3/s cubic foot per second,cubic meter per 2.831685E 02ft3/ssecond, m3/s foot to the fourthmeter to the fourth 8.630975E 038 power, ft4 (area power, m4moment of inertia) foot per minute,meter per second, 5.080000 E 03ft/min m/s foot per second,meter per second, 3.048000 E 01ft/sm/s TLFeBOOK 26. 40816_01_p1-1410/22/0112:38 PM Page 8 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 8CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of MeasurementT (Continued)( To convert from ToMultiply by foot per secondmeter per second 3.048000 E 01 h squared, ft/s2 squared, m/s2 h footcandle, fc lux, lx1.076391E 01 footlambert, fLcandela per square 3.426259E 00 hmeter, cd/m2 foot pound force, ftlbf joule, J 1.355818E 00 h foot pound force per watt, W2.259697E 02minute, ftlbf/minh foot pound force per watt, W1.355818E 00second, ftlbf/sh foot poundal, ft joule, J 4.214011E 02 hpoundal h free fall, standard gmeter per second 9.806650 E 00 isquared, m/s2 i gallon, gal (Canadiancubic meter, m34.546090E 03 liquid)i gallon, gal (U.K.cubic meter, m34.546092E 03liquid) i gallon, gal (U.S. dry) cubic meter, m34.404884E 03 gallon, gal (U.S.cubic meter, m33.785412E 03liquid) s gallon, gal (U.S.cubic meter per4.381264E 08 cliquid) per daysecond, m3/s gallon, gal (U.S.cubic meter per6.309020E 05liquid) per minute second, m3/s i grad degree (angular) 9.000000 E 01 grad radian, rad1.570796E 02 grain, grkilogram, kg 6.479891 E 05 i gram, gkilogram, kg 1.000000 E 03 TLFeBOOK 27. 40816_01_p1-1410/22/01 12:38 PM Page 9 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001CONVERSION FACTORS 9 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from ToMultiply by1hectare, hasquare meter, m 21.000000 E 04 horsepower, hp watt, W 7.456999 E 021 (550 ftlbf/s)0horsepower, hp watt, W 9.80950 E 03(boiler)0horsepower, hp watt, W 7.460000 E 022 (electric) horsepower, hp watt, W 7.46043E 020 (water) horsepower, hp (U.K.)watt, W 7.4570E 022hour, hsecond, s 3.600000 E 03 hour, h (sidereal) second, s 3.590170E 030inch, in meter, m2.540000 E 02 inch of mercury, in Hg pascal, Pa3.38638 E 033 (32F) (pressure) inch of mercury, in Hg pascal, Pa3.37685 E 033 (60F) (pressure) inch of water, inpascal, Pa2.4884E 023 H2O (60F)3 (pressure) square inch, in2 square meter, m26.451600 E 048cubic inch, in3cubic meter, m3 1.638706 E 05(volume or section5 modulus) inch to the fourth meter to the fourth 4.162314E 071 power, in4 (areapower, m42 moment of inertia)5inch per second, in/smeter per second, 2.540000 E 023 m/sTLFeBOOK 28. 40816_01_p1-1410/22/0112:38 PM Page 10 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 10 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of MeasurementT (Continued)(To convert fromToMultiply by kelvin, Kdegree Celsius, C tC TK 273.15 m kilogram force, kgfnewton, N9.806650 E 00 m kilogram force meter,newton meter,9.806650 E 00 mkgm Nmm kilogram force secondkilogram, kg 9.806650 E 00 msquared per meter,mkgfs2/m (mass) kilogram force per pascal, Pa 9.806650 E 04 msquare centimeter,skgf/cm2 kilogram force per pascal, Pa 9.806650 E 00 ssquare meter,kgf/m2m kilogram force per pascal, Pa 9.806650 E 06square millimeter,mkgf/mm2 kilometer per hour,meter per second,2.777778 E 01mkm/hm/s m kilowatt hour, kWh joule, J 3.600000 E 06 kip (1000 lbf) newton, N4.448222 E 03m kipper square inch,pascal, Pa 6.894757 E 06mkip/in2 ksi m knot, kn (international) meter per second,5.144444 E 01om/s lambert, L candela per square 3.183099 E 03ometer, cd/m litercubic meter, m31.000000 E 03 o maxwellweber, Wb1.000000 E 08 o mhosiemens, S 1.000000 E 00 o TLFeBOOK 29. 40816_01_p1-1410/22/01 12:38 PMPage 11 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001CONVERSION FACTORS11 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by microinch, in meter, m2.540000 E 080micron, m meter, m1.000000 E 060mil, mimeter, m2.540000 E 05 mile, mi (international) meter, m1.609344 E 030mile, mi (U.S. statute)meter, m1.609347E 03 mile, mi (internationalmeter, m1.852000 E 03 nautical)4mile, mi (U.S. nautical) meter, m1.852000 E 03 square mile, mi2 square meter, m22.589988 E 06 (international)0square mile, mi2 square meter, m22.589998 E 06 (U.S. statute) mile per hour, mi/hmeter per second, 4.470400 E 016(international) m/s mile per hour, mi/hkilometer per hour, 1.609344 E 00 (international) km/h1millibar, mbar pascal, Pa1.000000 E 02 millimeter of mercury, pascal, Pa1.33322 E 026mmHg (0C)3minute (angle) radian, rad 2.908882E 046minute, minsecond, s 6.000000 E 01 minute (sidereal)second, s 5.983617E 011ounce, ozkilogram, kg2.834952E 02 (avoirdupois)3ounce, oz (troy or kilogram, kg3.110348 E 02 apothecary)3ounce, oz (U.K. uid)cubic meter, m3 2.841307 E 058ounce, oz (U.S. uid)cubic meter, m3 2.957353 E 050ounce force, ozf newton, N 2.780139 E 01TLFeBOOK 30. 40816_01_p1-1410/22/01 12:38 PM Page 12 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 12 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement T (Continued) ( To convert from To Multiply by ounce forceinch,newton meter, 7.061552 E 03p ozfinNm ounce per square foot, kilogram per square 3.051517 E 01p oz (avoirdupois)/ft2meter, kg/m2 ounce per square yard, kilogram per square 3.390575 E 02p oz (avoirdupois)/yd2meter, kg/m2 perm (0C) kilogram per pascal 5.72135 E 11 p second meter, kg/(Pasm) p perm (23C)kilogram per pascal 5.74525 E 11 second meter, p kg/(Pasm) perm inch, permin kilogram per pascal 1.45322 E 12 p (0C) second meter, kg/(Pasm) p perm inch, permin kilogram per pascal 1.45929 E 12 (23C)second meter, p kg/(Pasm) pint, pt (U.S. dry)cubic meter, m3 5.506105 E 04p pint, pt (U.S. liquid) cubic meter, m3 4.731765 E 04 poise, p (absolute pascal second,1.000000 E 01 p viscosity)Pasp pound, lbkilogram, kg4.535924 E 01p (avoirdupois) pound, lb (troy or kilogram, kg3.732417 E 01p apothecary) pound square inch, kilogram square 2.926397 E 04p lbin2 (moment of meter, kgm2 inertia)pTLFeBOOK 31. 40816_01_p1-1410/22/01 12:38 PM Page 13 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001CONVERSION FACTORS13 TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from To Multiply by pound per footpascal second,1.488164 E 00 second, lb/fts Pas pound per square kilogram per square 4.882428 E 00 foot, lb/ft2meter, kg/m22pound per cubickilogram per cubic1.601846 E 01 foot, lb/ft3meter, kg/m3 pound per gallon,kilogram per cubic9.977633 E 01 lb/gal (U.K. liquid)meter, kg/m3 pound per gallon,kilogram per cubic1.198264 E 021lb/gal (U.S. liquid)meter, kg/m3 pound per hour, lb/h kilogram per1.259979 E 04 second, kg/s2pound per cubic inch,kilogram per cubic2.767990 E 04 lb/in3meter, kg/m3 pound per minute,kilogram per7.559873 E 032lb/minsecond, kg/s pound per second,kilogram per4.535924 E 01 lb/ssecond, kg/s4pound per cubic yard,kilogram per cubic5.932764 E 014lb/yd3meter, kg/m31poundalnewton, N 1.382550 E 01 poundforce, lbf newton, N 4.448222 E 001pound force foot,newton meter, 1.355818 E 00 lbfftNm1pound force per foot,newton per meter, 1.459390 E 01 lbf/ftN/m4pound force perpascal, Pa4.788026 E 01 square foot, lbf/ft2 pound force per inch,newton per meter, 1.751268 E 02 lbf/in N/mTLFeBOOK 32. 40816_01_p1-14 10/22/01 12:38 PMPage 14 40816 HICKS Mcghp Chap_01 Second Pass 7/3/2001 14 CHAPTER ONE TABLE 1.3 Factors for Conversion to SI Units of Measurement (Continued) To convert from ToMultiply by pound force per squarepascal, Pa6.894757 E 03inch, lbf/in2 (psi) quart, qt (U.S. dry)cubic meter, m3 1.101221E 03 quart, qt (U.S. liquid) cubic meter, m3 9.463529E 04 rod meter, m5.029210E 00 second (angle)radian, rad 4.848137E 06 second (sidereal) second, s 9.972696E 01 square (100 ft2)square meter, m29.290304 E 00 ton (assay) kilogram, kg2.916667E 02 ton (long, 2240 lb) kilogram, kg1.016047E 03 ton (metric)kilogram, kg1.000000 E 03 ton (refrigeration) watt, W 3.516800E 03 ton (register)cubic meter, m3 2.831685E 00 ton (short, 2000 lb)kilogram, kg9.071847E 02 ton (long per cubic kilogram per cubic1.328939E 03 yard, ton)/yd3 meter, kg/m3 ton (short per cubickilogram per cubic1.186553 E 03 yard, ton)/yd3 meter, kg/m3 ton force (2000 lbf)newton, N 8.896444 E 03 tonne, tkilogram, kg1.000000 E 03 watt hour, Wh joule, J3.600000 E 03 yard, ydmeter, m9.144000 E 01 square yard, yd2square meter, m28.361274 E 01 cubic yard, yd3 cubic meter, m3 7.645549 E 01 year (365 days) second, s 3.153600 E 07 year (sidereal) second, s 3.155815 E 07 Exact value. From E380, Standard for Metric Practice, American Society for Testing and Materials. TLFeBOOK 33. 40816_02_p15-5110/22/01 12:39 PMPage 15 40816 HICKS Mcghp Chap_02 First Pass 5/23/01 p. 15 CHAPTER 2 BEAM FORMULAS TLFeBOOKCopyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 34. 40816_02_p15-5110/22/0112:39 PM Page 1640816 HICKS Mcghp Ch02 Second Passpb 7/3/01 p. 16 408 16CHAPTER TWO In analyzing beams of various types, the geometric properties of a variety of cross-sectional areas are used. Figure 2.1 gives equations for computing area A, moment of inertia I, section modulus or the ratio S I/c, where c distance from the neutral axis to the outermost ber of the beam or other member. Units used are inches and millimeters and their powers. The formulas in Fig. 2.1 are valid for both USCS and SI units. Handy formulas for some dozen different types of beams are given in Fig. 2.2. In Fig. 2.2, both USCS and SI units can be used in any of the formulas that are applicable to both steel and wooden beams. Note that W load, lb (kN); L length, ft (m); R reaction, lb (kN); V shear, lb (kN); M bending moment, lb ft (N m); D deection, ft (m); a spacing, ft (m); b spacing, ft (m); E modulus of elasticity, lb/in2 (kPa); I moment of inertia, in4 (dm4); less than; greater than. Figure 2.3 gives the elastic-curve equations for a variety of prismatic beams. In these equations the load is given as P, lb (kN). Spacing is given as k, ft (m) and c, ft (m). CONTINUOUS BEAMS Continuous beams and frames are statically indeterminate. Bending moments in these beams are functions of the geometry, moments of inertia, loads, spans, and modulus of elasticity of individual members. Figure 2.4 shows how any span of a continuous beam can be treated as a single beam, with the moment diagram decomposed into basic components. Formulas for analysis are given in the diagram. Reactions of a continuous beam can be found by using the formulas in Fig. 2.5. Fixed-end moment formulas for beams of constant moment of inertia (prismatic beams) for TLFeBOOK 35. 40816_02_p15-51 10/22/01 12:39 PM Page 17 1640816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 17TLFeBOOK-1 ,erdh f Ieb , - ,ys FIGURE 2.1 Geometric properties of sections. .efy ,- .err17 36. 40816_02_p15-51 10/22/01 12:39 PM Page 1840816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 18408TLFeBOOK 18 37. 40816_02_p15-51 10/22/01 12:39 PM Page 191840816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 19TLFeBOOKGeometric properties of sections.FIGURE 2.1 (Continued) 19 38. 40816_02_p15-51 10/22/01 12:39 PM Page 2040816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 20408TLFeBOOK 20 39. 40816_02_p15-51 10/22/01 12:39 PM Page 212040816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 21 TLFeBOOK Geometric properties of sections. FIGURE 2.1 (Continued) 21 40. 40816_02_p15-51 10/22/01 12:39 PM Page 2240816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 22408TLFeBOOK 22 41. 40816_02_p15-51 10/22/01 12:39 PM Page 232240816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 23TLFeBOOKGeometric properties of sections.FIGURE 2.1 (Continued) 23 42. 40816_02_p15-51 10/22/01 12:39 PM Page 2440816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 24408TLFeBOOK 24 43. 40816_02_p15-51 10/22/01 12:39 PM Page 252440816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 25TLFeBOOKGeometric properties of sections.FIGURE 2.1 (Continued) 25 44. 40816_02_p15-51 10/22/01 12:39 PM Page 2640816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 26408TLFeBOOK s C i m i i b g m b d FGeometric properties of sections. b e e b l l m o d s FFIGURE 2.1 (Continued) f l t 26 45. 40816_02_p15-51 10/22/01 12:39 PM Page 272640816 HICKS Mcghp Ch02 Third Pass bcj 7/19/01 p. 27BEAM FORMULAS 27several common types of loading are given in Fig. 2.6.Curves (Fig. 2.7) can be used to speed computation ofxed-end moments in prismatic beams. Before the curvesin Fig. 2.7 can be used, the characteristics of the loadingmust be computed by using the formulas in Fig. 2.8. Theseinclude xL, the location of the center of gravity of the load-ing with respect to one of the loads; G2 b 2 Pn/W, wherenbnL is the distance from each load Pn to the center ofgravity of the loading (taken positive to the right); and S 3 b 3 Pn/W. These values are given in Fig. 2.8 for some com-nmon types of loading.Formulas for moments due to deection of a xed-endbeam are given in Fig. 2.9. To use the modied momentdistribution method for a xed-end beam such as that inFig. 2.9, we must rst know the xed-end moments for abeam with supports at different levels. In Fig. 2.9, the rightend of a beam with span L is at a height d above the leftend. To nd the xed-end moments, we rst deect thebeam with both ends hinged; and then x the right end,leaving the left end hinged, as in Fig. 2.9b. By noting that ap pline connecting the two supports makes an angle approxi-mately equal to d/L (its tangent) with the original positionof the beam, we apply a moment at the hinged end to pro-duce an end rotation there equal to d/L. By the denition ofstiffness, this moment equals that shown at the left end ofFig. 2.9b. The carryover to the right end is shown as the topformula on the right-hand side of Fig. 2.9b. By using thelaw of reciprocal deections, we obtain the end moments ofthe deected beam in Fig. 2.9 as d M F K F (1 C F) L LR (2.1) L d M F K F (1 C F) R RL (2.2) LTLFeBOOK 46. 40816_02_p15-51 10/22/01 12:39 PM Page 2840816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 28408TLFeBOOK 28 47. 28 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/01 12:39 PM29Third Pass Page 29bcj 7/19/01 p. 29 FIGURE 2.2 Beam formulas. (From J. Callender, Time-Saver Standards for Architectural Design Data, 6th ed., McGraw-Hill, N.Y.) TLFeBOOK 48. 40816_02_p15-51 10/22/01 12:39 PM Page 3040816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 30408TLFeBOOK 30 49. 40816_02_p15-51 10/22/01 12:39 PM Page 313040816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 31TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 31 50. 40816_02_p15-51 10/22/01 12:39 PM Page 3240816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 32408TLFeBOOK 32 51. 40816_02_p15-51 10/22/01 12:39 PM Page 333240816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 33TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 33 52. 40816_02_p15-51 10/22/01 12:39 PM Page 3440816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 34408TLFeBOOK 34 53. 40816_02_p15-51 10/22/01 12:39 PM Page 353440816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 35TLFeBOOK Beam formulas. FIGURE 2.2 (Continued) 35 54. 40816_02_p15-51 10/22/01 12:39 PM Page 3640816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 36408TLFeBOOK 36 55. 40816_02_p15-51 10/22/01 12:39 PM Page 373640816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 37 TLFeBOOKBeam formulas.FIGURE 2.2 (Continued) 37 56. 40816_02_p15-51 10/22/01 12:39 PM Page 3840816 HICKS Mcghp Ch02Second Pass pb 7/3/01 p. 38408TLFeBOOK 38 57. 40816_02_p15-51 10/22/01 12:39 PM Page 393840816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 39 TLFeBOOKBeam formulas.FIGURE 2.2 (Continued) 39 58. 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/0140 12:39 PMSecond Pass Page 40pb 7/3/01 p. 40 FIGURE 2.3 Elastic-curve equations for prismatic beams. (a) Shears, moments, deections for full uniform load on a simply supported prismatic beam. (b) Shears and moments for uniform load over part of a simply supported prismatic beam. (c) Shears, moments, deections for a concentrated load at any point of a simply supported prismatic beam. TLFeBOOK408 59. 4040816_02_p15-51 40816 HICKS Mcghp Ch0210/22/0112:39 PM41 Third PassPage 41 bcj 7/19/01 p. 41 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (d) Shears, moments, deections for a concentrated load at midspan of a simply supported prismatic beam. (e) Shears, moments, deections for two equal concentrated loads on a simply supported prismatic beam. ( f ) Shears, moments, deections for several equal loads equally spaced on a simply supported prismatic beam.TLFeBOOK 60. 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/0142 12:39 PMSecond Pass Page 42pb 7/3/01 p. 42 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (g) Shears, moments, deections for a concentrated load on a beam overhang. (h) Shears, moments, deections for a concentrated load on the end of a prismatic cantilever. (i) Shears, moments, deections for a uniform load over the full length of a beam with overhang. TLFeBOOK408 61. 4240816_02_p15-51 40816 HICKS Mcghp Ch0210/22/0112:39 PM43 Third PassPage 43 bcj 7/19/01 p. 43 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (j) Shears, moments, deections for uniform load over the full length of a cantilever. (k) Shears, moments, deections for uniform load on a beam overhang. (l) Shears, moments, deections for triangular loading on a prismatic cantilever.TLFeBOOK 62. 40816_02_p15-51 40816 HICKS Mcghp Ch0210/22/014412:39 PM Second PassPage 44 pb 7/3/01 p. 44 FIGURE 2.3 (Continued)Elastic-curve equations for prismatic beams. (m) Simple beamload increasing uniformly to one end.TLFeBOOK 408 63. 44 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/01 12:39 PM45Third Pass Page 45bcj 7/19/01 p. 45 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (n) Simple beamload increasing uniformly to center. TLFeBOOK 64. 40816_02_p15-51 40816 HICKS Mcghp Ch0210/22/014612:39 PM Second PassPage 46 pb 7/3/01 p. 46 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (o) Simple beamuniform load partially distributed at one end.TLFeBOOK 408 65. 46 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/01 12:39 PM47Third Pass Page 47bcj 7/19/01 p. 47 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (p) Cantilever beamconcen- trated load at free end. TLFeBOOK 66. 40816_02_p15-5140816 HICKS Mcghp Ch02 10/22/0148 12:39 PMSecond Pass Page 48pb 7/3/01 p. 48 FIGURE 2.3 (Continued) Elastic-curve equations for prismatic beams. (q) Beam xed at both ends concentrated load at center. TLFeBOOK408 67. 4840816_02_p15-51 40816 HICKS Mcghp Ch0210/22/0112:39 PM49 Third PassPage 49 bcj 7/19/01 p. 49 FIGURE 2.4 Any span of a continuous beam (a) can be treated as a simple beam, as shown in (b) and (c). In (c), the moment diagram is decomposed into basic components.TLFeBOOK 68. 40816_02_p15-51 10/22/01 12:39 PMPage 5040816 HICKS Mcghp Ch02 Second Pass pb 7/3/01 p. 50408 50 CHAPTER TWOFccIof FIGURE 2.5 Reactions of continuous beam (a) found by making the beam statically determinate. (b) Deections computed with interior supports removed. (c), (d ), and (e) Deections calculated for unit load over each removed support, to obtain equations for w each redundant.it TLFeBOOK 69. 40816_02_p15-51 10/22/0112:39 PMPage 515040816 HICKS Mcghp Ch02Third Passbcj 7/19/01 p. 51 BEAM FORMULAS51FIGURE 2.6 Fixed-end moments for a prismatic beam. (a) For aconcentrated load. (b) For a uniform load. (c) For two equal con-centrated loads. (d) For three equal concentrated loads.In a similar manner the xed-end moment for a beam withone end hinged and the supports at different levels can befound fromdg MF K (2.3)h Ldr where K is the actual stiffness for the end of the beam thatis xed; for beams of variable moment of inertia K equalsthe xed-end stiffness times (1 C FC F). L R TLFeBOOK 70. 40816_02_p52-9810/22/0112:41 PM Page 52 40816 HICKS Mcghp Ch02Third Pass 7/19/01 bcj p. 52 408 FIGURE 2.7 Chart for xed-end moments due to any type of loading.FCMbuFIGURE 2.8 Characteristics of loadings. TLFeBOOK 71. 40816_02_p52-9810/22/01 12:41 PMPage 53p. 52 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 53 BEAM FORMULAS53f FIGURE 2.8 (Continued)Characteristics of loadings. ULTIMATE STRENGTH OF CONTINUOUS BEAMS Methods for computing the ultimate strength of continuous beams and frames may be based on two theorems that x upper and lower limits for load-carrying capacity: 1. Upper-bound theorem. A load computed on the basis ofan assumed link mechanism is always greater than, or atbest equal to, the ultimate load.TLFeBOOK 72. 40816_02_p52-9810/22/01 12:41 PM Page 54 40816 HICKS Mcghp Ch02 Third Pass7/19/01 bcj p. 54408 54 CHAPTER TWOptmttrfUtiFIGURE 2.9 Moments due to deection of a xed-end beam.C2. Lower-bound theorem. The load corresponding to ano equilibrium condition with arbitrarily assumed values for the redundants is smaller than, or at best equal to, the ultimate loadingprovided that everywhere moments do not exceed MP. The equilibrium method, based on the lower bound theorem, is usually easier for simpleA cases. kTLFeBOOK 73. 40816_02_p52-98 10/22/01 12:41 PMPage 55p. 54 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 55 BEAM FORMULAS55 For the continuous beam in Fig. 2.10, the ratio of the plastic moment for the end spans is k times that for the center span (k 1). Figure 2.10b shows the moment diagram for the beam made determinate by ignoring the moments at B and C and the moment diagram for end moments MB and MC applied to the determinate beam. Then, by using Fig. 2.10c, equilibrium is maintained when wL2 1 1 MP M M42 B 2 C wL2 4 kM P wL2 (2.4) 4(1 k) The mechanism method can be used to analyze rigid frames of constant section with xed bases, as in Fig. 2.11. Using this method with the vertical load at midspan equal to 1.5 times the lateral load, the ultimate load for the frame is 4.8MP /L laterally and 7.2M P /L vertically at midspan. Maximum moment occurs in the interior spans AB and CD when L Mx (2.5) 2 wLnor ifse L kM Ps M kM P when x (2.6)2wLneA plastic hinge forms at this point when the moment equals kMP. For equilibrium,TLFeBOOK 74. 40816_02_p52-98 10/22/0112:41 PMPage 56 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 56 408 56CHAPTER TWO FIGURE 2.10 Continuous beam shown in (a) carries twice as much uniform load in the center span as in the side span. In (b) are shown the moment diagrams for this loading condition with redundants removed and for the redundants. The two moment diagrams are combined in (c), producing peaks at which plastic hinges are assumed to form.TLFeBOOK 75. p. 56 e e s s -40816_02_p52-98 40816 HICKS Mcghp Ch0210/22/0112:41 PM57 Second PassPage 57 7/3/01 pb p. 57 FIGURE 2.11 Ultimate-load possibilities for a rigid frame of constant section with xed bases.TLFeBOOK 76. 40816_02_p52-9810/22/0112:41 PMPage 58 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 58408 58 CHAPTER TWOw xkM P x (L x) kM P2 Lw2 L kM L KM 1 kM kM2wLP 2wL P2 wL P 2P leading to F k 2M 2P wL2 3kM P 0wL 2 4 (2.7) When the value of MP previously computed is substituted,s o7k 2 4k 4ork (k 47) 47 o s from which k 0.523. The ultimate load is s 4M P (1 k)M fwL 6.1 P(2.8) LLm iIn any continuous beam, the bending moment at any section is equal to the bending moment at any other section, plus the shear at that section times its arm, plus the product of all the intervening external forces times their respective arms. Thus, in Fig. 2.12, Vx R 1 R 2 R 3 P1 P2 P3 M x R 1 (l 1 l 2 x) R 2 (l 2 x) R 3x P1 (l 2 c x) P2 (b x) P3a M x M 3 V3x P3aTable 2.1 gives the value of the moment at the various F supports of a uniformly loaded continuous beam over equal a TLFeBOOK 77. 40816_02_p52-98 10/22/0112:41 PM Page 59p. 58 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 59 BEAM FORMULAS59 FIGURE 2.12 Continuous beam.) , spans, and it also gives the values of the shears on each side of the supports. Note that the shear is of the opposite sign on either side of the supports and that the sum of the two shears is equal to the reaction. Figure 2.13 shows the relation between the moment and shear diagrams for a uniformly loaded continuous beam of)four equal spans. (See Table 2.1.) Table 2.1 also gives the maximum bending moment that occurs between supports, in addition to the position of this moment and the points ofy ,tesFIGURE 2.13 Relation between moment and shear diagrams forla uniformly loaded continuous beam of four equal spans.TLFeBOOK 78. 40816_02_p52-9840816 HICKS Mcghp Ch02 TABLE 2.1 Uniformly Loaded Continuous Beams over Equal Spans (Uniform load per unit length w; length of each span l)Distance to 10/22/01 point of Distance toShear on each side max moment,point ofNotation of support. L left, measured to inection,Number of R right. Reaction atMoment Maxright frommeasured toofsupport any support is L Rover eachmoment in fromright from60 12:41 PM supports of span L Rsupport each spansupportsupportThird Pass2 1 or 20 1 200.125 0.500None3 103 800.07030.3750.7502 5 8 5 8 1 8 0.07030.6250.25010 Page 60 44 1 0 0 0.080 0.4000.8007/19/01 bcj p. 602 106510110 0.025 0.500 0.276, 0.7241 011 280 0.07720.3930.7865 2172815 28 328 0.03640.536 0.266, 0.8063132813 28 228 0.03640.464 0.194, 0.7341 015 380 0.07790.3950.789 TLFeBOOK4086 22320 4 0 03320 526 0 268 0 783 79. 38 p. 6040816_02_p52-98 40816 HICKS Mcghp Ch02622338 2038438 0.0332 0.526 0.268, 0.783 31838 1938338 0.0461 0.500 0.196, 0.804 1041 10400.0777 0.3940.788 2 63 10455 104 111040.0340 0.533 0.268, 0.79073 49 10451 104 8 104 0.0433 0.490 0.196, 0.78510/22/01 4 53 10453 104 9 104 0.0433 0.510 0.215, 0.804 1 0 56 142 0 0.0778 0.3940.78982 86 14275 142 151420.0338 0.528 0.268, 0.788 3 67 14270 142 111420.0440 0.493 0.196, 0.790 4 72 14271 142 121420.0405 0.500 0.215, 0.78512:41 PM wlwl wl 2wl 2 ll61 Values apply to Second Pass The numerical values given are coefcients of the expressions at the foot of each column.Page 61 7/3/01 pb p. 61TLFeBOOK 80. 40816_02_p52-9810/22/01 12:41 PM Page 62 40816 HICKS Mcghp Ch02 Third Pass7/19/01 bcj p. 62408 62CHAPTER TWOtrtrtw FIGURE 2.14 Values of the functions for a uniformly loaded B continuous beam resting on three equal spans with four supports.B inection. Figure 2.14 shows the values of the functions for s a uniformly loaded continuous beam resting on three equall spans with four supports.Mt Maxwells Theoremwa When a number of loads rest upon a beam, the deection ati any point is equal to the sum of the deections at this point due to each of the loads taken separately. Maxwells theo- t rem states that if unit loads rest upon a beam at two points,s A and B, the deection at A due to the unit load at B equals t the deection at B due to the unit load at A.ei Castiglianos Theoremm2 This theorem states that the deection of the point of s application of an external force acting on a beam is equal aTLFeBOOK 81. 40816_02_p52-98 10/22/01 12:41 PM Page 63p. 62 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 63BEAM FORMULAS 63 to the partial derivative of the work of deformation with respect to this force. Thus, if P is the force, f is the deection, and U is the work of deformation, which equals the resilience:dU fdP According to the principle of least work, the deformation of any structure takes place in such a manner that the work of deformation is a minimum.dBEAMS OF UNIFORM STRENGTH . Beams of uniform strength so vary in section that the unitrstress S remains constant, and I/c varies as M. For rectangu-llar beams of breadth b and depth d, I/c I/c bd 2/6 and M Sbd2/6. For a cantilever beam of rectangular cross section, under a load P, Px Sbd 2/6. If b is constant, d 2 varies with x, and the prole of the shape of the beam is a parabola, as in Fig. 2.15. If d is constant, b varies as x, and the beamtis triangular in plan (Fig. 2.16).tShear at the end of a beam necessitates modication of-the forms determined earlier. The area required to resist,shear is P/Sv in a cantilever and R/Sv in a simple beam. Dot-sted extensions in Figs. 2.15 and 2.16 show the changes necessary to enable these cantilevers to resist shear. The waste in material and extra cost in fabricating, however, make many of the forms impractical, except for cast iron. Figure 2.17 shows some of the simple sections of uniformfstrength. In none of these, however, is shear taken intolaccount.TLFeBOOK 82. 40816_02_p52-98 10/22/0112:41 PM Page 64 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 64 408 64 CHAPTER TWOFIGURE 2.15 Parabolic beam of uniform strength. SAFE LOADS FOR BEAMS OF VARIOUS TYPES Table 2.2 gives 32 formulas for computing the approximate safe loads on steel beams of various cross sections for an T allowable stress of 16,000 lb/in2 (110.3 MPa). Use these W formulas for quick estimation of the safe load for any steel p beam you are using in a design.lTLFeBOOK 83. 40816_02_p52-98 10/22/0112:41 PMPage 65p. 64 40816 HICKS Mcghp Ch02 Second Pass 7/3/01 pb p. 65BEAM FORMULAS 65FIGURE 2.16 Triangular beam of uniform strength.e Table 2.3 gives coefcients for correcting values innTable 2.2 for various methods of support and loading.eWhen combined with Table 2.2, the two sets of formulaslprovide useful time-saving means of making quick safe- load computations in both the ofce and the eld. TLFeBOOK 84. 40816_02_p52-98 40816 HICKS Mcghp Ch0210/22/01 TABLE 2.2 Approximate Safe Loads in Pounds (kgf) on Steel Beams* (Beams supported at both ends; allowable ber stress for steel, 16,000 lb/in2 (1.127 kgf/cm2) (basis of table) for iron, reduce values given in table by one-eighth)Greatest safe load, lb Deection, in6612:41 PM Shape of sectionLoad in middle Load distributed Load in middleLoad distributed Third Pass 890AD1,780AD wL3wL3 Solid rectangle LL32AD 2 52AD 2 890(AD ad) 1,780(AD ad) wL3 wL3Page 66 Hollow rectangle 7/19/01 bcj p. 66LL32(AD 2 ad 2) 52(AD 2 ad 2) 667AD1,333AD wL3wL3 Solid cylinder LL24AD 2 38AD 2 667(AD ad) 1,333(AD ad) wL3 wL3 Hollow cylinderLL24(AD 2 ad 2) 38(AD 2 ad 2)TLFeBOOK 408 885AD1 770AD L3 L3 85. L L 24(AD ad )38(AD ad ) p. 6640816_02_p52-98 40816 HICKS Mcghp Ch02885AD1.770AD wL3 wL3 Even-legged angle or teeLL32AD 252AD 2 1,525AD 3,050AD wL3 wL3 Channel or Z bar L L53AD 285AD 210/22/01 1,380AD 2,760AD wL3 wL3 Deck beam L L50AD 280AD 2 1,795AD 3,390AD wL3 wL3 I beam L L58AD 293AD 212:41 PM *L distance between supports, ft (m); A sectional area of beam, in2 (cm2); D depth of beam, in (cm);67 a interior area, in2 (cm2); d interior depth, in (cm); w total working load, net tons (kgf). Second PassPage 67 7/3/01 pb p. 67TLFeBOOK 86. 40816_02_p52-98 40816 HICKS Mcghp Ch0210/22/01 TABLE 2.3 Coefcients for Correcting Values in Table 2.2 for Various Methods of Support and of Loading6812:41 PM Max relative Third Passdeection under Max relative max relative safe Conditions of loadingsafe load load Beam supported at endsPage 68 Load uniformly distributed over span 1.0 1.0 7/19/01 bcj p. 68 Load concentrated at center of span 120.80 Two equal loads symmetrically concentratedl/4c Load increasing uniformly to one end0.9740.976 Load increasing uniformly to center 340.96 Load decreasing uniformly to center 321.08TLFeBOOK 408 B d d il 87. g y 2p. 68 40816_02_p52-9840816 HICKS Mcghp Ch02 Beam xed at one end, cantilever Load uniformly distributed over span 1 4 2.40 Load concentrated at end 1 8 3.20 Load increasing uniformly to xed end3 8 1.92 Beam continuous over two supports equidistant from ends 10/22/01 Load uniformly distributed over span 1. If distance a 0.2071ll2/4a2 2. If distance a 0.2071l l/(l 4a) 3. If distance a 0.2071l5.83 Two equal loads concentrated at endsl/4a 12:41 PMl length of beam; c distance from support to nearest concentrated load; a distance from support to end69 Second Pass of beam. Page 697/3/01 pb p. 69 TLFeBOOK 88. 40816_02_p52-98 10/22/01 12:41 PMPage 70 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 70408 TLFeBOOK 70 89. 40816_02_p52-98 10/22/01 12:41 PM Page 71p. 70 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 71 TLFeBOOK FIGURE 2.17 Beams of uniform strength (in bending). 71 90. 40816_02_p52-98 10/22/01 12:41 PMPage 72 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 72408 TLFeBOOK 72 91. 40816_02_p52-98 10/22/01 12:41 PM Page 73p. 72 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 73 TLFeBOOK Beams of uniform strength (in bending). FIGURE 2.17 (Continued) 73 92. 40816_02_p52-98 10/22/01 12:41 PMPage 74 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 74408 TLFeBOOK 74 93. 40816_02_p52-98 10/22/01 12:41 PM Page 75p. 74 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 75TLFeBOOKBeams of uniform strength (in bending).FIGURE 2.17 (Continued) 75 94. 40816_02_p52-98 10/22/01 12:41 PMPage 76 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 76408 TLFeBOOK 76 95. 40816_02_p52-98 10/22/01 12:41 PM Page 77p. 76 40816 HICKS Mcghp Ch02Second Pass7/3/01 pb p. 77 TLFeBOOK Beams of uniform strength (in bending). FIGURE 2.17 (Continued) 77 96. 40816_02_p52-98 10/22/01 12:41 PMPage 78 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 78 408TLFeBOOKRRpwwbsmBeams of uniform strength (in bending).FIGURE 2.17 (Continued)ab 78 97. 40816_02_p52-9810/22/01 12:41 PM Page 79p. 78 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 79BEAM FORMULAS79 ROLLING AND MOVING LOADS Rolling and moving loads are loads that may change their position on a beam or beams. Figure 2.18 shows a beam with two equal concentrated moving loads, such as two wheels on a crane girder, or the wheels of a truck on a bridge. Because the maximum moment occurs where the shear is zero, the shear diagram shows that the maximum moment occurs under a wheel. Thus, with x a/2:R1 P 1 2x l a l M2 Pl2 1al2x a l l4x 2 2 lR2 P 1 2x l a l Pl1 al 2a 2 2x 3a4x 2 M1 22l 2l ll M2 max when x 14a M1 max when x 34a Pl1 2l 2l l 2 aP a 2 2 M max 2Figure 2.19 shows the condition when two equal loads are equally distant on opposite sides of the center of the beam. The moment is then equal under the two loads. TLFeBOOK 98. 40816_02_p52-98 10/22/01 12:41 PM Page 80 40816 HICKS Mcghp Ch02Third Pass 7/19/01 bcj p. 80 408 t r b t s b t p FIGURE 2.18 Two equal concentrated moving loads.FIGURE 2.19 Two equal moving loads equally distant on oppo-site sides of the center.F80 TLFeBOOK 99. 40816_02_p52-98 10/22/0112:41 PMPage 81p. 80 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 81BEAM FORMULAS81 If two moving loads are of unequal weight, the condition for maximum moment is the maximum moment occurring under the heavier wheel when the center of the beam bisects the distance between the resultant of the loads and the heavier wheel. Figure 2.20 shows this position and the shear and moment diagrams. When several wheel loads constituting a system are on a beam or beams, the several wheels must be examined in turn to determine which causes the greatest moment. The position for the greatest moment that can occur under a given - FIGURE 2.20Two moving loads of unequal weight. TLFeBOOK 100. 40816_02_p52-98 10/22/0112:41 PM Page 82 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 82408 82 CHAPTER TWO wheel is, as stated earlier, when the center of the span bisects the distance between the wheel in question and the resultant of all loads then on the span. The position for maximum shear at the support is when one wheel is passing off the span. CURVED BEAMS The application of the flexure formula for a straight beam to the case of a curved beam results in error. When all fibers of a member have the same center of curvature, the concentric or common type of curved beam exists (Fig. 2.21). Such a beam is defined by the WinklerBach theory. The stress at a point y units from the centroidal axis is SMAR 1 Z (Ry y) Z M is the bending moment, positive when it increases curva-f ture; y is positive when measured toward the convex side; A is the cross-sectional area; R is the radius of the centroidali axis; Z is a cross-section property dened by Z 1AyRydA TABLE 2 4 A l i l EAnalytical expressions for Z of certain sections are given in Table 2.4. Z can also be found by graphical integration methods (see any advanced strength book). The neutral surface shifts toward the center of curvature, or inside ber, an amount equal to e ZR/(Z 1). The WinklerBach TLFeBOOK 101. mA p. 82 h n n hnes-f- l lll,,40816_02_p52-98 40816 HICKS Mcghp Ch02 TABLE 2.4 Analytical Expressions for ZSectionExpressionR ln R C RC10/22/01 Z 1 h R R R 1 2 Z 1 212:41 PMr rr83 Second Pass Z 1 RAt ln (R C1) (b t) ln (R C0) b ln (R C2) and A tC1 (b t) C3 bC2Page 83 7/3/01 pb p. 83 Z 1 RA b ln R C2 R C2 (t b) ln R C RC 1 1 A 2 [(t b)C1 bC2]TLFeBOOK 102. 40816_02_p52-9810/22/01 12:41 PM Page 84 40816 HICKS Mcghp Ch02 Third Pass7/19/01 bcj p. 84408 84CHAPTER TWOBFIGURE 2.21 Curved beam. theory, though practically satisfactory, disregards radial stresses as well as lateral deformations and assumes pureB bending. The maximum stress occurring on the inside ber is S Mhi /AeRi, whereas that on the outside ber is S Mh0 /AeR0. The deection in curved beams can be computed by means of the moment-area theory. The resultant deection is then equal to 0 2 2x y in the direction dened by tan y / x. Deections can also be found conveniently by use of Castiglianos theorem. It states that in an elastic system the displacement in the direction of a force (or couple) and due to that force (or couple) is the partial derivative of the strain energy with respect to the force (or couple). A quadrant of radius R is xed at one end as shown in Fig. 2.22. The force F is applied in the radial direction at free-end B. Then, the deection of B isFTLFeBOOK 103. 40816_02_p52-9810/22/01 12:41 PM Page 85p. 84 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 85BEAM FORMULAS85 By moment area, y R sin x R(1 cos ) ds Rd M FR sin FR 3 FR 3 Bx By 4EI2EIFR3 2and B 2EI 14at x tan 1 FR 32EI 4EI FR 3 2 tan 1 32.5 leBy Castigliano, r FR3FR3 Bx By 4EI2EIy 2 yn .e-tnt FIGURE 2.22Quadrant with xed end. TLFeBOOK 104. 40816_02_p52-98 10/22/01 12:41 PM Page 86 40816 HICKS Mcghp Ch02Third Pass7/19/01 bcj p. 864 86 CHAPTER TWOEccentrically Curved BeamsT(These beams (Fig. 2.23) are bounded by arcs having differ-ent centers of curvature. In addition, it is possible for either1radius to be the larger one. The one in which the sectiondepth shortens as the central section is approached may becalled the arch beam. When the central section is thelargest, the beam is of the crescent type.Crescent I denotes the beam of larger outside radius andcrescent II of larger inside radius. The stress at the centralsection of such beams may be found from S KMC/I.In the case of rectangular cross section, the equation2becomes S 6KM/bh2, where M is the bending moment, bis the width of the beam section, and h its height. The stressfactors, K for the inner boundary, established from photo-elastic data, are given in Table 2.5. The outside radius3iccFIGURE 2.23 Eccentrically curved beams. a TLFeBOOK 105. 40816_02_p52-9810/22/0112:41 PM Page 87p. 8640816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 87BEAM FORMULAS87 TABLE 2.5 Stress Factors for Inner Boundary at Central Section (see Fig. 2.23)-r1. For the arch-type beamsnehRo Ri(a) K 0.834 1.504if 5e Ro Rih hR Rid (b) K 0.899 1.181 if 5 o 10 Ro Ri h l(c) In the case of larger section ratios use the equivalent beamI.solutionn2. For the crescent I-type beamsb s hRo Ri(a) K 0.570 1.536if 2 -Ro Rih s h Ro Ri(b) K 0.959 0.769if2 20Ro Ri h R R h Ro Ri 0.0298(c) K 1.092 if 20o ih 3. For the crescent II-type beams hRo Ri(a) K 0.897 1.098if 8Ro Rih R R hRo Ri0.0378(b) K 1.119 if8 20o i h h Ro Ri 0.0270(c) K 1.081 if 20 Ro Ri h is denoted by Ro and the inside by Ri. The geometry of crescent beams is such that the stress can be larger in off- center sections. The stress at the central section determined above must then be multiplied by the position factor k, TLFeBOOK 106. 40816_02_p52-98 10/22/0112:41 PMPage 88 40816 HICKS Mcghp Ch02Third Pass7/19/01 bcj p. 88408 88CHAPTER TWO given in Table 2.6. As in the concentric beam, the neutral surface shifts slightly toward the inner boundary. (See Vidosic, Curved Beams with Eccentric Boundaries, Transactions of the ASME, 79, pp. 13171321.) w s ELASTIC LATERAL BUCKLING OF BEAMS m f When lateral buckling of a beam occurs, the beam under- goes a combination of twist and out-of-plane bending (Fig. 2.24). For a simply supported beam of rectangular T cross section subjected to uniform bending, buckling occurs at the critical bending moment, given by( AMcr EIyGJ L where L unbraced length of the member E modulus of elasticity Iy moment of inertial about minor axisG shear modulus of elasticity J torsional constant The critical moment is proportional to both the lateral bending stiffness EIy /L and the torsional stiffness of the member GJ/L. For the case of an open section, such as a wide-ange or I-beam section, warping rigidity can provide additional torsional stiffness. Buckling of a simply supported beam of open cross section subjected to uniform bending occurs at theb critical bending moment, given by b TLFeBOOK 107. 40816_02_p52-9810/22/0112:41 PM Page 89p. 88 40816 HICKS Mcghp Ch02Second Pass 7/3/01 pb p. 89BEAM FORMULAS89le Mcr L EIy GJ ECw 2 L2 where Cw is the warping constant, a function of cross- sectional shape and dimensions (Fig. 2.25).In the preceding equations, the distribution of bending moment is assumed to be uniform. For the case of a nonuniform bending-moment gradient, buckling often occurs at-grTABLE 2.6 Crescent-Beam Position Stress Factorss (see Fig. 2.23)k Angle ,degreeInner Outer 10 1 0.055 H/h 1 0.03 H/h 20 1 0.164 H/h 1 0.10 H/h 30 1 0.365 H / h 1 0.25 H/h 40 1 0.567 H / h 1 0.467 H/h (0.5171 1.382 H/h)1/2 50 1.521 1 0.733 H/h1.382 (0.2416 0.6506 H/h)1/2 60 1.756 1 1.123 H/h0.6506 (0.4817 1.298 H/h)1/2 70 2.070 1 1.70 H/hl0.6492e(0.2939 0.7084 H/h)1/2 80 2.531 1 2.383 H/h0.3542r-90 1 3.933 H/hnNote: All formulas are valid for 0 H/h 0.325. Formulas for the innereboundary, except for 40 degrees, may be used to H/h 0.36. H distance between centers. TLFeBOOK 108. 40816_02_p52-98 10/22/0112:41 PM Page 90 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 90 408 90 CHAPTER TWOFAIYFIGURE 2.24 (a) Simple beam subjected to equal end moments.(b) Elastic lateral buckling of the beam.a a larger critical moment. Approximation of this critical bending moment, Mcr may be obtained by multiplying Mcr given by the previous equations by an amplication factorM cr Cb Mcr12.5Mmax where Cb 2.5Mmax 3MA 4MB 3MCTLFeBOOK 109. 40816_02_p52-98 10/22/01 12:41 PM Page 91p. 90 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 91 BEAM FORMULAS91 FIGURE 2.25 Torsion-bending constants for torsional buckling. A cross-sectional area; Ix moment of inertia about xx axis; Iy moment of inertia about yy axis. (After McGraw-Hill, New York). Bleich, F., Buckling Strength of Metal Structures.s. and Mmax absolute value of maximum moment in theunbraced beam segmentlMA absolute value of moment at quarter point ofr the unbraced beam segmentr MB absolute value of moment at centerline of theunbraced beam segment MC absolute value of moment at three-quarterpoint of the unbraced beam segment TLFeBOOK 110. 40816_02_p52-9810/22/01 12:41 PM Page 92 40816 HICKS Mcghp Ch02 Third Pass7/19/01 bcj p. 92408 92CHAPTER TWOCb equals 1.0 for unbraced cantilevers and for members where the moment within a signicant portion of the unbraced segment is greater than, or equal to, the larger of the segment end moments. wsao COMBINED AXIAL AND BENDING LOADS a For short beams, subjected to both transverse and axial loads, the stresses are given by the principle of superposition if the deection due to bending may be neglected without serious error. That is, the total stress is given with sufcient w accuracy at any section by the sum of the axial stress and ( the bending stresses. The maximum stress, lb/in2 (MPa), equalsUP Mc f AIW where P axial load, lb (N )cmA cross-sectional area, in2 (mm2) ta M maximum bending moment, in lb (Nm) ppc distance from neutral axis to outermost ber atthe section where maximum moment occurs,in (mm)I moment of inertia about neutral axis at thatsection, in4 (mm4)w When the deection due to bending is large and the axial load produces bending stresses that cannot be neglected, the maximum stress is given byTLFeBOOK 111. 40816_02_p52-98 10/22/0112:41 PMPage 93p. 92 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 93 BEAM FORMULAS 93s Pcef (M Pd)AIf where d is the deection of the beam. For axial compression, the moment Pd should be given the same sign as M; and for tension, the opposite sign, but the minimum value of M Pd is zero. The deection d for axial compression and bending can be closely approximated byldodn 1 (P/Pc)ttwhere do deection for the transverse loading alone, ind(mm); and Pc critical buckling load 2EI / L2, lb (N). , UNSYMMETRICAL BENDING When a beam is subjected to loads that do not lie in a plane containing a principal axis of each cross section, unsymmetrical bending occurs. Assuming that the bending axis of the beam lies in the plane of the loads, to preclude torsion, and that the loads are perpendicular to the bending axis, to preclude axial components, the stress, lb/in2 (MPa), at any point in a cross section ist,Mx yM x f y IxIyt whereMx bending moment about principal axis XX, in lb (Nm)e- My bending moment about principal axis YY, in lb (Nm) TLFeBOOK 112. 40816_02_p52-98 10/22/01 12:41 PM Page 94 40816 HICKS Mcghp Ch02Third Pass7/19/01 bcj p. 94 408 94 CHAPTER TWOx distance from point where stress is to becomputed to YY axis, in (mm)y distance from point to XX axis, in (mm) Ix moment of inertia of cross section about XX,Fin (mm4)cIy moment of inertia about YY, in (mm4) asIf the plane of the loads makes an angle with a principal f plane, the neutral surface forms an angle with the other n principal plane such thatsD Ix tan tan Iy tsmt ECCENTRIC LOADING If an eccentric longitudinal load is applied to a bar in the P plane of symmetry, it produces a bending moment Pe, where e is the distance, in (mm), of the load P from the o centroidal axis. The total unit stress is the sum of thisa moment and the stress due to P applied as an axial load: (f P A PecI P A1 ec r2 where A cross-sectional area, in2 (mm2)w c distance from neutral axis to outermost ber, in (mm)TLFeBOOK 113. 40816_02_p52-98 10/22/01 12:41 PM Page 95p. 94 40816 HICKS Mcghp Ch02 Second Pass7/3/01 pb p. 95BEAM FORMULAS 95 I moment of inertia of cross section about neutral axis, in4 (mm4) r radius of gyration I/A, in (mm) Figure 2.1 gives values of the radius of gyration for several cross sections. If there is to be no tension on the cross section under a compressive load, e should not exceed r 2/c. For a rectangular section with width b, and depth d, the eccentricity, there-lfore, should be less than b/6 and d/6 (i.e., the load shouldrnot be applied outside the middle third). For a circular cross section with diameter D, the eccentricity should not exceed D/8. When the eccentric longitudinal load produces a deection too large to be neglected in computing the bending stress, account must be taken of the additional bending moment Pd, where d is the deection, in (mm). This deection may be closely approximated by4eP/Pc d(1 P/Pc)ePc is the critical buckling load 2EI/L2, lb (N).,If the load P, does not lie in a plane containing an axiseof symmetry, it produces bending about the two principalsaxes through the centroid of the section. The stresses, lb/in2 (MPa), are given by P Pexcx Peycyf AIyIx where A cross-sectional area, in2 (mm2)nex eccentricity with respect to principal axis YY,in (mm)TLFeBOOK 114. 40816_02_p52-98 10/22/0112:41 PM Page 96 40816 HICKS Mcghp Ch02 Third Pass 7/19/01 bcj p. 96408 96 CHAPTER TWO ey eccentricity with respect to principal axis XX,in (mm) cx distance from YY to outermost ber, in (mm) cy distance from XX to outermost ber, in (mm) Ix moment of inertia about XX, in4 (mm4) Iy moment of inertia about YY, in4 (mm4) The principal axes are the two perpendicular axes through the centroid for which the moments of inertia are a maximum or a minimum and for which the products of inertia are zero. NATURAL CIRCULAR FREQUENCIES AND NATURAL PERIODS OF VIBRATION OF PRISMATIC BEAMS Figure 2.26 shows the characteristic shape and gives constants for determination of natural circular frequency and natural period T, for the rst four modes of cantilever, simply supported, xed-end, and xed-hinged beams. To obtain , select the appropriate constant from Fig. 2.26 and multiply it by EI/wL4. To get T, divide the appropriate constant by EI/wL4.In these equations, natural frequency, rad/sW beam weight, lb per linear ft (kg per linear m) L beam length, ft (m) TLFeBOOK 115. p. 96dodhea-- )) ,,40816_02_p52-98 40816 HICKS Mcghp Ch0210/22/0112:41 PM97 Second PassPage 97 7/3/01 pb p. 97 FIGURE 2.26 Coefcients for computing natural circular frequencies and natural periods of vibration of prismatic beams.TLFeBOOK 116. 40816_02_p52-98 10/22/0112:41 PMPage 98 40816 HICKS Mcghp Ch02Third Pass 7/19/01 bcj p. 98 98CHAPTER TWO E modulus of elasticity, lb/in2 (MPa) I moment of inertia of beam cross section, in4 (mm4) T natural period, s To determine the characteristic shapes and natural periods for beams with variable cross section and mass, use the Rayleigh method. Convert the beam into a lumped-mass system by dividing the span into elements and assuming the mass of each element to be concentrated at its center. Also, compute all quantities, such as deection and bending moment, at the center of each element. Start with an assumed characteristic shape. ]TLFeBOOK 117. 40816_03_p99-130 10/22/01 12:42 PM Page 99 40816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 99 CHAPTER 3COLUMN FORMULAS TLFeBOOKCopyright 2002 The McGraw-Hill Companies. Click Here for Terms of Use. 118. 40816_03_p99-13010/22/0112:42 PMPage 10040816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 100 40816 100 CHAPTER THREE GENERAL CONSIDERATIONS Columns are structural members subjected to direct compression. All columns can be grouped into the following three classes: 1. Compression blocks are so short (with a slendernessratio that is, unsupported length divided by the leastradius of gyration of the member below 30) that bend-TABLE 3 1 Strength of Round-Ended Columns According to Eulers Formula*ing is not potentially occurring. 2. Columns so slender that bending under load is a given aretermed long columns and are dened by Eulers theory. 3. Intermediate-length columns, often used in structuralpractice, are called short columns.Long and short columns usually fail by buckling when their critical load is reached. Long columns are analyzed using Eulers column formula, namely,n 2EI n 2EAPcr l 2(l/r)2 In this formula, the coefcient n accounts for end conditions. When the column is pivoted at both ends, n 1; when one end is xed and the other end is rounded, n 2; when both ends are xed, n 4; and when one end is xed and the other is free, n 0.25. The slenderness ratio sepa- rating long columns from short columns depends on the modulus of elasticity and the yield strength of the column material. When Eulers formula results in (Pcr /A) > Sy, strength instead of buckling causes failure, and the column ceases to be long. In quick estimating numbers, this critical slenderness ratio falls between 120 and 150. Table 3.1 gives additional column data based on Eulers formula. TLFeBOOK 119. nnddnge ess----l;; lt , . p. 10040816_03_p99-130 40816 HICKS Mcghp Chap_03TABLE 3.1 Strength of Round-Ended Columns According to Eulers Formula* Low-Medium-Wrought carboncarbonMaterial Cast ironironsteel steel 2Ultimate compressive strength, lb/in 107,00053,40062,60089,00010/22/01Allowable compressive stress, lb/in2(maximum)7,10015,40017,00020,000Modulus of elasticity14,200,000 28,400,00030,600,00031,300,000Factor of safety 85 5 5Smallest I allowable at worst section, in4 Pl 2 Pl 2Pl 2Pl 212:42 PM101 17,500,000 56,000,00060,300,00061,700,000 SECOND PASS 7/5/01 pb p. 101Limit of ratio, l/r 50.0 60.659.455.6 Rectangle r b112 , l/b 14.4 17.517.216.0Circle r 14 d, l/d 12.5 15.214.913.9Page 101Circular ring of small thicknessr d , l/d 1817.6 21.421.119.7*(P allowable load, lb; l length of column, in; b smallest dimension of a rectangular section, in; d diameter of acircular section, in; r least radius of gyration of section.)To convert to SI units, use: 1b/in2 6.894 kPa; in4 (25.4)4 mm4.TLFeBOOK 120. 40816_03_p99-13010/22/01 12:42 PM Page 10240816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 10240816 102CHAPTER THREE FIGURE 3.1 L/r plot for columns. SHORT COLUMNS Stress in short columns can be considered to be partly due to compression and partly due to bending. Empirical, rational expressions for column stress are, in general, based on the assumption that the permissible stress must be reduced below that which could be permitted were it due to compression only. The manner in which this reduction is made determines the type of equation and the slenderness ratio beyond which the equation does not apply. Figure 3.1 shows the curves for this situation. Typical column formulas are given in Table 3.2. ECCENTRIC LOADS ON COLUMNS When short blocks are loaded eccentrically in compression or in tension, that is, not through the center of gravity (cg), a combination of axial and bending stress results. The maximum unit stress SM is the algebraic sum of these two unit stresses. TLFeBOOK 121. won h n oe e srl ,.p. 102 40816_03_p99-13040816 HICKS Mcghp Chap_03TABLE 3.2 Typical Short-Column Formulas Formula Material CodeSlenderness ratio 10/22/01 l l 2Sw 17,000 0.485Carbon steelsAISC 120 r rSw 16,000 70 rl Carbon steelsChicago l r 120S 15,000 50 l l103 12:42 PM Carbon steelsAREA 150SECOND PASS 7/5/01 pb p. 103w r rS 19,000 100 l lwCarbon steelsAm. Br. Co. 60 120 r rc r15.9 l 2 l S 135,000 65 Page 103cr Alloy-steel tubing ANC crS 9,000 40 l lwCast ironNYC 70 r r TLFeBOOK 122. 40816_03_p99-13040816 HICKS Mcghp Chap_03TABLE 3.2 Typical Short-Column Formulas (Continued) FormulaMaterial CodeSlenderness ratioScr 34,500 245 l1 94 10/22/01 2017ST aluminumANCc r cr rl 2 0.51Scr 5,000 Spruce ANC 72ccr104 Sy rl l 2n2E 2SECOND PASS 7/5/01 pb p. 104 12:42 PMScr Sy 1 Steels Johnson4n 2E r Sy SylScr Steels Secant critical rec1 2 secr l r P 4AE Page 104Scr theoretical maximum; c end xity coefcient; c 2, both ends pivoted; c 2.86, one pivoted, other xed;c 4, both ends xed; c 1 one xed, one free.is initial eccentricity at which load is applied to center of column cross section. TLFeBOOK40816ty a t a d F 123. 40816_03_p99-13010/22/01 12:42 PM Page 105p. 10440816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 105 COLUMN FORMULAS105FIGURE 3.2 Load plot for columns.In Fig. 3.2, a load, P, acts in a line of symmetry at thedistance e from cg; r radius of gyration. The unit stressesare (1) Sc, due to P, as if it acted through cg, and (2) Sb, dueto the bending moment of P acting with a leverage of eabout cg. Thus, unit stress, S, at any point y is S Sc Sb (P/A) Pey/I Sc(1 ey/r 2)y is positive for points on the same side of cg as P, and nega-tive on the opposite side. For a rectangular cross section of TLFeBOOK 124. 40816_03_p99-130 10/22/0112:42 PMPage 10640816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 106 40816 106 CHAPTER THREE width b, the maximum stress, SM Sc (1 6e/b). When P is outside the middle third of width b and is a compressivem load, tensile stresses occur.s For a circular cross section of diameter d, SM Sc(1 t 8e/d). The stress due to the weight of the solid modies t these relations. 3 Note that in these formulas e is measured from the grav- 0 ity axis and gives tension when e is greater than one-sixthf the width (measured in the same direction as e), for rectan- p gular sections, and when greater than one-eighth the diam- t eter, for solid circular sections. (a FIGURE 3.3 Load plot for columns.F TLFeBOOK 125. 40816_03_p99-13010/22/0112:42 PM Page 107 p. 10640816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 107COLUMN FORMULAS 107PIf, as in certain classes of masonry construction, theematerial cannot withstand tensile stress and, thus, no tension can occur, the center of moments (Fig. 3.3) is taken atthe center of stress. For a rectangular section, P acts at dis- s tance k from the nearest edge. Length under compression 3k, and SM 3P/hk. For a circular section, SM [0.372 2-0.056(k/r)]P/k rk , where r radius and k distance of Phfrom circumference. For a circular ring, S average com--pressive stress on cross section produced by P; e eccen--tricity of P; z length of diameter under compression (Fig. 3.4). Values of z/r and of the ratio of Smax to average S are given in Tables 3.3 and 3.4. FIGURE 3.4 Circular column load plot. TLFeBOOK 126. 40816_03_p99-130 40816 HICKS Mcghp Chap_03TABLE 3.3 Values of the Ratio z/r(See Fig. 3.5)r1e r er0.0 0.50.60.70.80.91.0 r10/22/010.25 2.000.250.30 1.820.300.35 1.661.89 1.98 0.350.40 1.511.75 1.84 1.930.401080.45 1.371.61 1.71 1.81 1.90 0.45 SECOND PASS 7/5/01 pb p. 10812:42 PM0.50 1.231.46 1.56 1.66 1.78 1.89 2.00 0.500.55 1.101.29 1.39 1.50 1.62 1.74 1.87 0.550.60 0.971.12 1.21 1.32 1.45 1.58 1.71 0.600.65 0.840.94 1.02 1.13 1.25 1.40 1.54 0.650.70 0.720.75 0.82 0.93 1.05 1.20 1.35 0.70Page 1080.75 0.590.60 0.64 0.72 0.85 0.99 1.15 0.750.80 0.470.47 0.48 0.52 0.61 0.77 0.94 0.800.85 0.350.35 0.35 0.36 0.42 0.55 0.72 0.850.90 0.240.24 0.24 0.24 0.24 0.32 0.49 0.900.95 0.120.12 0.12 0.12 0.12 0.12 0.25 0.95TLFeBOOK 40816 127. p. 108 40816_03_p99-13040816 HICKS Mcghp Chap_03TABLE 3.4 Values of the Ratio Smax/Savg(In determining S average, use load P divided by total area of cross section) 10/22/01 r1 e r e r 0.00.50.6 0.70.8 0.91.0 r0.001.00 1.001.00 1.00 1.00 1.00 1.00 0.00109 12:42 PM0.051.20 1.161.15 1.13 1.12 1.11 1.10 0.05SECOND PASS 7/5/01 pb p. 1090.101.40 1.321.29 1.27 1.24 1.22 1.20 0.100.151.60 1.481.44 1.40 1.37 1.33 1.30 0.150.201.80 1.641.59 1.54 1.49 1.44 1.40 0.200.252.00 1.801.73 1.67 1.61 1.55 1.50 0.250.302.23 1.961.88 1.81 1.73 1.66 1.60 0.30 Page 1090.352.48 2.122.04 1.94 1.85 1.77 1.70 0.350.402.76 2.292.20 2.07 1.98 1.88 1.80 0.400.453.11 2.512.39 2.23 2.10 1.99 1.90 0.45 TLFeBOOK 128. 40816_03_p99-13010/22/0112:42 PM Page 11040816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 110 40816 110 CHAPTER THREE The kern is the area around the center of gravity of a cross section within which any load applied produces stress of only one sign throughout the entire cross section. Outside the kern, a load produces stresses of different sign. Figure 3.5 shows kerns (shaded) for various sections. c For a circular ring, the radius of the kern r 0 D[1(d/D)2]/8.w For a hollow square (H and h lengths of outer and inner sides), the kern is a square similar to Fig. 3.5a, where C B a t T u t w t FIGURE 3.5 Column characteristics.1TLFeBOOK 129. 40816_03_p99-130 10/22/0112:42 PM Page 111p. 11040816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 111 COLUMN FORMULAS1111 H 0.1179H 1 H f H 1 h2h 2srmin s6 2fsFor a hollow octagon, Ra and Ri radii of circles cir-cumscribing the outer and inner sides; thickness of wall 0.9239(Ra Ri); and the kern is an octagon similar to Fig. 3.5c,where 0.2256R becomes 0.2256Ra[1 (Ri /Ra)2].deCOLUMN BASE PLATE DESIGNBase plates are usually used to distribute column loads overa large enough area of supporting concrete construction thatthe design bearing strength of the concrete is not exceeded.The factored load, Pu, is considered to be uniformly distrib-uted under a base plate.The nominal bearing strength fp kip/in2 or ksi (MPa) ofthe concrete is given by A1 A2fp 0.85f c A1 and A1 2where f specied compressive strength of concrete,cksi (MPa)A1 area of the base plate, in2 (mm2)A2 area of the supporting concrete that is geo- metrically similar to and concentric with the loaded area, in2 (mm2) In most cases, the bearing strength, fp is 0.85f c , whenthe concrete support is slightly larger than the base plate or 1.7f c , when the support is a spread footing, pile cap, or matTLFeBOOK 130. 40816_03_p99-13010/22/0112:42 PMPage 11240816 HICKS Mcghp Chap_03 SECOND PASS 7/5/01 pb p. 11240816 112CHAPTER THREE foundation. Therefore, the required area of a base plate for a factored load Pu is Pu A1 c0.85 fc where c is the strength reduction factor 0.6. For a wide- ange column, A1 should not be less than bf d, where bf is the ange width, in (mm), and d is the depth of column, in (mm). The length N, in (mm), of a rectangular base plate for a wide-ange column may be taken in the direction of d as N A1 d or 0.5(0.95d 0.80bf) AC The width B, in (mm), parallel to the anges, then, isD A1 B NTa The thickness of the base plate tp, in (mm), is the largestm of the values given by the equations that follow:2Putp m 0.9Fy BN 2Puwtp n 0.9Fy BNp2Pu it p n0.9Fy BNct wherem projection of base plate beyond the ange sand parallel to the web, in (mm)ti (N 0.95d)/2 c TLFeBOOK 131. 40816_03_p99-13010/22/0112:42 PMPage 113p. 11240816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 113 COLUMN FORMULAS 113rn projection of base plate beyond the edges of the ange and perpendicular to the web, in (mm) (B 0.80bf)/2n (dbf)/4- (2X)/[1 (1 X)] 1.0s,X [(4 dbf)/(d bf)2][Pu/( 0.85f 1)]caAMERICAN INSTITUTE OF STEELCONSTRUCTION ALLOWABLE-STRESSDESIGN APPROACHThe lowest columns of a structure usually are supported ona concrete foundation. The area, in square inches (squaret millimeters), required is found from: PA FPwhere P is the load, kip (N) and Fp is the allowable bearingpressure on support, ksi (MPa).The allowable pressure depends on strength of concretein the foundation and relative sizes of base plate and con-crete support area. If the base plate occupies the full area ofthe support, Fp 0.35f c , where f c is the 28-day compres-sive strength of the concrete. If the base plate covers lessthan the full area, FP 0.35fc A2/A1 0.70f c , where A1 is the base-plate area (B N), and A2 is the full area of theconcrete support.TLFeBOOK 132. 40816_03_p99-130 10/22/01 12:42 PMPage 11440816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 114 40816 114 CHAPTER THREE Eccentricity of loading or presence of bending momentT at the column base increases the pressure on some partss of the base plate and decreases it on other parts. To com- b pute these effects, the base plate may be assumed com- b pletely rigid so that the pressure variation on the concrete t is linear. t Plate thickness may be determined by treating projections m and n of the base plate beyond the column as cantilevers.wcpstCTfdt FIGURE 3.6 Column welded to a base plate.c TLFeBOOK 133. 40816_03_p99-13010/22/0112:42 PM Page 115p. 11440816 HICKS Mcghp Chap_03SECOND PASS 7/5/01 pb p. 115 COLUMN FORMULAS 115t The cantilever dimensions m and n are usually dened ass shown in Fig. 3.6. (If the base plate is small, the area of the- base plate inside the column profile should be treated as a- beam.) Yield-line analysis shows that an equivalent can-e tilever dimension n can be defined as n 14dbf , andthe required base plate thickness tp can be calculated froms.fptp 2l Fywhere l max (m, n, n), in (mm)fp P/(BN) Fp, ksi (MPa) Fy yield strength of base plate, ksi (MPa)P column axial load, kip (N) For columns subjected only to direct load, the welds ofcolumn to base plate, as shown in Fig. 3.6, are requiredprincipally for withstanding erection stresses. For columnssubjected to uplift, the welds must be proportioned to resistthe forces.COMPOSITE COLUMNSThe AISC load-and-resistance factor design (LRFD) speci-fication for struct

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